# Tag Info

0

This seems to be a design with repeated measures within subjects. This means that measurements within a particular subject are likely to be more similar to each other than to other subjects. That is, there will be correlation within subjects. One way to proceed with this is to use a mixed effects model and fit random intercepts for subject. In the common ...

2

As also pointed out in the comments, you don't need prior since all you need is the posterior: $$P(\text{H_1 is true}|\mathbf{x})=P(\theta\leq0.5|\mathbf{x})=\int_0^{0.5} \pi(\theta|\mathbf{x})d\theta$$ Since this is Beta distribution, $0\leq\theta\leq1$, a uniform prior on $\theta$ would be $\pi(\theta)=1$ and you wouldn't notice it in the integration ...

1

Both an ANOVA and mixed modeling approach will work for your data. I am more familiar with mixed models, and can speak better to that. However, you are correct that you can run your analysis as a mixed model and then get an ANOVA table afterward. If you are using R, then you can use lmer to first estimate the mixed model: require(lme4) m <- lmer(outcome ~...

2

The geometric distribution is a special case of the negative binomial distribution, so I would try a negative binomial glm (generalized linear model.) Some details here: When to use Poisson vs. geometric vs. negative binomial GLMs for count data?. The generalization of ANOVA in glm's is called the analysis of deviance, but in some computer systems (for ...

0

No, you conduct the experiment only once. Carrying forward your example, suppose you are conducting a binomial experiment which consists of 1000 Bernoulli trials. Then, your null hypothesis might be that the probability of success, i.e., q = 0.5. If this is actually true, then you would see around 500 heads and 500 tails. However, having too many heads ...

0

Yes, for N stationary time series, or for N residuals from which the serial correlation has been filtered, it's necessary to apply N univariate SPRTs in parallel for anomaly detection.

2

Many standard statistical tests can be reframed as simple linear models. So in your example, there is no difference between a two sample T-test testing for differences in X and Y versus a linear regression of Y and X against a grouping indicator. Your intuition that regression requires more data is not correct. Also, as whuber mentions in his comment, you ...

2

By definition, 'white noise' is serially uncorrelated. It is like hitting piano keys at random; note n+1 cannot at all be predicted by note n. This is distinguished from brown noise and pink noise, both of which have some serial correlation, or dependence, albeit to different extents. To answer your question, 'strict white noise' assume essentially 0 serial ...

2

STATISTICAL SIGNIFICANCE $\ne$ PRACTICAL SIGNIFIANCE You have a pretty large sample size, meaning that you have the power to reject a null hypothesis of equal means, since standard error is a function of both effect size and sample size. You don't dispute that $7.9 \ne 8$, do you? Consequently, if you have that one distribution has a mean of $7.9$ and ...

1

This effect is a form of publication bias, but also an effect of wholesale adoption of significance at P $\leq$ 0.05, ignoring effect size, and the crisis of reproducibility in science. I add the alpha level of 0.05, as this level is not particularly stringent for the reason the OP states - 5/100 studies will be statistically significant by chance alone. ...

4

This scenario seems like a straightforward example of publication bias. The definition at the linked Wikpedia article is a good one: publication bias "occurs when the outcome of an experiment or research study influences the decision whether to publish or otherwise distribute it." In your scenario, the 95 experiments that failed to reject the null hypothesis ...

1

A "probability distribution" can be uniquely described either by its CDF or the corresponding probability measure. Contrarily, a density function is often not a unique description of a probability distribution, and so it would not usually be used for statements of the equality of distributions.$^\dagger$ Unfortunately, there is no standard notation for ...

0

See https://www.mathworks.com/matlabcentral/fileexchange/60147-normality-test-package. This package automatically runs 10 goodness of fit tests: normalitytest(X) Make sure X is a row vector. This function provides ten Normality tests that are not altogether available under one compact routine as a compiled Matlab function. All tests are coded to ...

4

If you want to provide evidence for equivalence (by contrast with evidence for difference), you can perform a t-test for equivalence using TOST. (In the below "$\theta$" is the difference between groups you are estimating a la $\bar{\mu}_{1} - \bar{\mu}_{2}$ or whatever.) General 'negativist' null hypothesis: $H_{0}^{-}: |\theta| \ge \delta$, with $H_{\text{... 0 Your research question is whether user experience differs between android apps and progressive web apps. You designed an experiment and collected data. Your results turned out not significant at the a priori chosen$\alpha$level of$0.05$. Assuming the data collection and all statistical assumptions were met, the p-value gives you the probability of ... 1 I will give some thoughts on all three issues I've raised. 100 covariates An issue here is that you will need a lot of data to get reliable estimates of the parameters in your regression. Otherwise, your model can badly overfit, meaning that you're basically guessing when you do inference on the group parameter. A solution is to collect more data or to be ... 1 The sample size term is not missing. It is implicit in the expression (Variance of the sample mean). This variance is equal to the variance of X divided by sample size, as the latter expression shows. With comparison of two sample means, it is possible that sample sizes differ. That is why there are two sample size terms in the latter case. 7 However this solution doesn't take into account the two replicates for each visit or separate visits. Correct. How would I do this? You need to account for repeated visits for each patient, and for repeated replicates within each visit for each patient. This is because measurements for the same patient are likely to be be more similar to measurements ... 1 The problem with Frequent Pattern Mining is that, in practice, one is likely to discover a titanic number of patterns with many redundancies. Many researchers have devoted substantial effort to come up with clever ways to improve the quality of pattern discovery, i.e. bring forth the most interesting patterns and bring down redundancy. The word '... 0 Consider$Y_i \sim N(\mu_i, \sigma^2)$(independently) as a random vector with a multivariate normal distribution,$\vec Y \sim N(\vec\mu, \sigma^2 I)$. The main idea is to consider this distribution with reference to a new coordinate system. We choose the first axis to be along$\vec 1 := (1, ..., 1)$. The projection of any vector$\vec y$onto this axis ... 4 This is six questions--and @StubbornAtom has done a good job outlining answers in one pithy comment. There appear to be two obstacles to going further: one is doing the initial algebra and the other is conceptual. Let's deal with the algebra first, because that's straightforward and doing it doesn't deprive you of the joy of discovery and learning the ... 1 Yes, if that is your alternative hypothesis: I have data in which I have a clinical suspicion that the mean of x is greater than the mean of y then you can use a one-tailed test. If it is going to be part of a paper for publication then you should perhaps consult the journal's guidelines since it might require a two-sided test. In any case, just be very ... 1 When the$t$-tests are performed, they assume that the other variables are already in the model. For example, suppose you were building a model where the dependent variable was the weight of a book, and the independent variables were$x_2$(the number of pages in the book) and$x_3$(the thickness of the book). If you fit a model with both of these ... 2 1) The density of your uniform in (-2,2) is 1/4, not 1. This has an impact on your$T$and$k$. 2) You can forget about$\gamma$, because both distributions are continuous, so$P_0(T=k)=0$. 3) This is somewhat nonstandard because$T$indicates that$h_1$is better outside$(-2,2)$,$h_0$is better in a set of the shape$(-2,-a)\cup(a,2)$, and$h_1$is ... 0 Let's write down a model matrix the way that we would for ANOVA. The difference is that the response variable will be binary instead of continuous, but that has no impact on the model matrix. As is typical, let's include of column of$1s$for the intercept and let$0$s in the rest of the columns represent some baseline group. (I will pick the group labeled 1,... 1 To say that the mean of a variable in one group differs from the mean of the variable in another group is to say that what group you're in has a relationship with the variable. For example, if men and women differed in average income, you would say there is a relationship between income and gender. So a test that compares the mean of the two groups is a ... 1 As @whuber notes in a comment, you do need to deal first with what seem to be incorrect premises in your approach. Most important, if your samples aren't truly random then "you can't use any of these methods to make inferences," as he put it. Fix that first. In terms of mean versus median as a measure of central tendency, the choice is yours based on your ... 0 Assuming, that the sequence of items is not important and has not been rearranged after one of the selection processes (see the comment of whuber), this sounds like a classical application of the Wald-Wolfowitz runs test, see Runs Test for Detecting Nonrandomness or Wald-Wolfowitz runs test. For example, in your case with$N_+=20$and$N_-=80$one gets a ... 1 First of all, I would suggest presenting confidence intervals instead of$p$-values. That said, I think the presented$z$values are a bit off. Using$Z = \frac{(p_b - p_a)}{ \sqrt{p_p (1-p_p) (\frac{1}{n_a} + \frac{1}{n_b})}}$, where$p_p$is the pooled proportion and$p_a$and$p_b$are the proportions observed in samples$a$and$b$respectively, we get$...

1

Yes, you can do Tukey on a within subjects variable. You'll find the info in Keppel's book on ANOVA. SPSS doesn't include tukey for within subjects variables, but you can easily compute it by hand.

0

Putting the three time points as three times as many cases does not require assuming independence. Identically replicating three times the measurements obtained at a single time point would essentially produce the same factors as the factor analysis on that single time point. Therefore independence is not a problem. Presenting the three time points as three ...

2

You are tackling the problem the wrong way. First you have a hypothesis. Was your hypothesis was that all questions were related to your effect of interest? If that's true, then the level of significance for this test is the maximum p-value you found. Therefore it is not significant. If your question was that at least one question was related to the ...

4

The Dickey-Fuller test addresses the question of whether the time series of interest has a unit root. The Ljung-Box and the Durbin-Watson tests help assess whether the time series of interest is autocorrelated. These are two different questions. According to the tag description of a unit root, A unit root is a property of a non-stationary time series ...

0

T1 will be more efficient if it has a small variance compered to T2

8

You appear to be using a Beta(1,1) prior on $\theta$. Since this is a continuous distribution, the prior (and posterior) probability of the event that the coin is exactly fair, $\theta=1/2$, is zero. What would perhaps be a more sensible prior (see Lindley 1957 pp. 188-189 for a discussion of similar examples) would be a point mass at $\theta=1/2$ given ...

1

First of all, a graph represents the trends in your sample but not in your population. Also, I assume that the number of the controls is greater than the number of idividuals with ASD. Thus they have greater variance and there could be some influencial points that make the difference. You should first find if there are any influential points in your ...

1

The statistical community's responses to the problem tend to assume that the answer lies in statistics. (The applied research community's preferred response is to ignore the problem entirely.) In a forthcoming comment, colleagues and I argue that purely statistical standard error underestimates uncertainty, and that behavioral researchers should commit to ...

1

There is a collection of alternatives in the special issue Statistical Inference in the 21st Century: A World Beyond p < 0.05 of The American Statistician that, I think, deserves special mention. I cannot possibly and won't try to list all the ideas about alternatives and/or additions to p-values that are given in the 43 papers of this special issue, but ...

2

$H_0$: Amount of missed shot is geometrically distributed with $p = 0.7$ $H_1$: Amount of missed shot is not geometrically distributed with $p = 0.7$ \begin{array} {|r|r|}\hline \text{amount of missed shots} & 0 & 1 & 2 & >2 \\ \hline \text{empirical frequency} & 72 & 18 & 8 & 2 \\ \hline \text{expected frequency} &...

8

No, that isn't valid. For instance, suppose you are interested in the relative heights of adult men and women in different parts of the world. You could find the usual difference across a broad population (say it's 9 cm, just for a guess). Then you look at (say) a tribe in New Guinea. Your null could be "Men in New Guinea are 9 cm taller than women". The ...

0

If you want to compare two samples, you should not be doing a one sample test. A one sample test where you take the first sample as if they were population figures leads the test to have a higher actual type I error rate than you chose. However, with time series you also have the additional issue that the usual assumption of independence is unlikely to ...

1

If there is no autocorrelation within each data set, then this could be handled by a mixed model. As the nature of each row is evidently the same in all the data sets, the 2000 rows would be included as random effects (e.g., the row number could serve the same function as a subject ID in other applications of mixed models) and the group and data set (...

1

Rather than expressing this in terms of the proportions, it is better to deal directly with the underlying count data. You essentially have a $2 \times 2 \times 2$ contingency table of count values, so you should be looking at tests of partial independence for that kind of data. Since you have small count values, it is probably feasible to use a Fisher ...

1

Hypothesis Testing - is drug effective or not? In Hypothesis testing usually the objective is to establish whether a treatment is effective or not. This can mean a number of things such as, whether a drug cures a disease or not? Whether offering discounts improve conversion rates or not? By convention Null Hypothesis $H_{0}$ assumes state of status-quo. ...

0

Math is your friend: $P_{null}$(Make at least one type I error) =$P_{null}$(any p-values less than the critical) =$P_{null}$(min p-values less than the critical) =$F_{minP}$(critical) Clearly you can see why you need to compare your p-values with the critical value from the min P distribution. I'll not be surprised if the minP and sidak correction ...

1

I like @NeilFultz answer, and I more than encourage you to learn, practise and expand your knowledge. However, if time is of the essence for you, I will provide the following solution. I guess if you are not completely comfortable with going beyond OLS-type questions, how about a practical solution of choosing a dataset, in which you can model a continuous ...

2

You probably want to use non-parametric methods to handle your ordinal data - for example, you could calculate the Spearman correlation (instead of Pearson), and test if the correlation between free time and Walc is zero or not. There are also nonparametric analogues of the t-test, ANOVA and others - many of them have examples in R on Choosing the correct ...

0

UMP test is not unique here. Since the pdf $f$ has monotone likelihood ratio (MLR) in $X_{(n)}$, by Karlin-Rubin theorem a UMP size $\alpha$ test for testing $H_0:\theta=\theta_0$ against $H_1:\theta<\theta_0$ is $$\phi_0(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(n)}<\theta_0\alpha^{1/n} \\ 0&,\text{ otherwise }\end{cases}$$ Now ...

0

The likelihood ratio test (LRT) statistic does not always equal $1$. Likelihood function given the sample $x=(x_1,\ldots,x_n)\in\{1,2,\ldots,N\}$ is $$L(N\mid x)=\frac{1}{N^n}I_{\{x_{(n)},x_{(n)}+1,\ldots\}}(N)\quad,\,N\in\{1,2,\ldots\}$$ Here $x_{(n)}=\max\{x_1,x_2,\ldots,x_n\}$ is the maximum order statistic as usual. So the LRT statistic is \begin{...

1

It's because the second part is for multivariate analysis in $p$ dimensions, unlike the first screenshot which deals with all scalars. It proceeds with an analogy from univariate case. In multivariate case, you won't be able to compare a statistics of the form $T=n(\bar{X}-\mu_0)S^{-1}$ against a threshold because it is $p$ dimensional. However, by analogy, ...

Top 50 recent answers are included